To prove this, note that \(Z^{\prime }\)  is normal, as \(Z^{\prime }\subset Z \times _Y Y^{\prime }\rightarrow Z\)  is finite étale and \(Z\)  is normal. Moreover, the morphism \(Z^{\prime }\rightarrow X^{\prime }\times Y^{\prime }\)  is finite and surjective. By [1]}, the Stein factorization of \(Z^{\prime }\rightarrow X\)  is étale over \(U\subset X\) . By [2]}, over the étale locus the Stein factorization commutes with taking fibres, so the Stein factorization of \(Z^{\prime }_x\rightarrow \operatorname{Spec}k(x)\)  is given by \(\operatorname{Spec}\Gamma (Z^{\prime }_x, \mathcal {O}_{Z^{\prime }_x})\) .
Since \(Z^{\prime }_{x} = Z_{x} \times _Y Y^{\prime }\)  is a disjoint union of copies of \(Y^{\prime }\)  (as it follows from Remark REF  and a standard normalization argument, noticing that \(Y^{\prime }\rightarrow Z_{x} \rightarrow Y\)  is the Galois closure of \(Z_x \rightarrow Y\) ), we see that \(Z^{\prime }_x \rightarrow X^{\prime }_x \times Y^{\prime }\)  is an isomorphism (as it is a finite surjective morphism between the same number of copies of \(Y^{\prime }\) ). It follows that \(Z^{\prime }\rightarrow X^{\prime }\times Y^{\prime }\)  is an isomorphism over a dense open subset, hence it is a birational morphism. Since \(Z^{\prime }\)  and \(X^{\prime }\times Y^{\prime }\)  are integral normal varieties over \(k\) , it follows as claimed that \(Z^{\prime }\rightarrow X^{\prime }\times Y^{\prime }\)  is an isomorphism by Zariski's Main Theorem (see [3]}).
